Differentiation Techniques: Chain Rule and Composite Functions
Composite Functions: Definition and Examples
In mathematics, new functions can be formed by combining existing functions in various ways. One powerful method of combining functions is called composition. A composite function arises when we apply one function to the output of another function.
Definition of a Composite Function
Let $f$ and $g$ be two functions. The composite function of $f$ with $g$, denoted by $(f \circ g)$, is defined by:
$(f \circ g)(x) = f(g(x))$
... (i)
This means that for a given input $x$, we first evaluate the "inner" function $g$ at $x$ to get $g(x)$. Then, we use this result, $g(x)$, as the input for the "outer" function $f$ to get $f(g(x))$.
The function $g$ is applied first, and then the function $f$ is applied to the result of $g$.
Similarly, the composite function of $g$ with $f$, denoted by $(g \circ f)$, is defined by:
$(g \circ f)(x) = g(f(x))$
... (ii)
In this case, the function $f$ is applied first, and then $g$ is applied to the result of $f$.
In general, $(f \circ g)(x) \neq (g \circ f)(x)$. Function composition is generally not commutative.
The domain of the composite function $(f \circ g)(x) = f(g(x))$ consists of all values of $x$ in the domain of $g$ such that the value $g(x)$ is in the domain of $f$. In other words, $x$ must be an allowable input for $g$, and the output $g(x)$ must be an allowable input for $f$.
Examples of Composite Functions
Let's look at several examples to understand how composite functions are formed and evaluated. Recognizing a function as a composition of simpler functions is crucial for applying the Chain Rule of differentiation.
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Let $f(u) = \sin u$ and $g(x) = x^2$.
Find $(f \circ g)(x)$ and $(g \circ f)(x)$.
$(f \circ g)(x) = f(g(x))$: Here, $g(x) = x^2$ is the inner function, and $f(u) = \sin u$ is the outer function.
$(f \circ g)(x) = f(x^2) = \sin(x^2)$
$(g \circ f)(x) = g(f(x))$: Here, $f(x) = \sin x$ is the inner function, and $g(v) = v^2$ is the outer function (using $v$ as the variable for $g$).
$(g \circ f)(x) = g(\sin x) = (\sin x)^2 = \sin^2 x$
Notice that $\sin(x^2)$ and $\sin^2 x$ are generally different functions, illustrating that composition is not commutative.
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Let $f(u) = \sqrt{u}$ and $g(x) = 3x + 5$.
Find $(f \circ g)(x)$ and its domain.
$(f \circ g)(x) = f(g(x))$: $g(x) = 3x+5$ (inner), $f(u) = \sqrt{u}$ (outer).
$(f \circ g)(x) = f(3x + 5) = \sqrt{3x + 5}$
Domain of $g(x) = 3x+5$: All real numbers, $(-\infty, \infty)$.
Domain of $f(u) = \sqrt{u}$: All non-negative numbers, $[0, \infty)$, i.e., $u \ge 0$.
For $f(g(x)) = \sqrt{3x+5}$ to be defined, $x$ must be in the domain of $g$, and $g(x)$ must be in the domain of $f$. So, $3x+5$ must be greater than or equal to 0.
$3x + 5 \ge 0$
$3x \ge -5$
$x \ge -\frac{5}{3}$
The domain of $(f \circ g)(x) = \sqrt{3x+5}$ is $\left[-\frac{5}{3}, \infty\right)$.
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Let $f(u) = u^5$ and $g(x) = \cos x$.
Find $(f \circ g)(x)$.
$(f \circ g)(x) = f(g(x))$: $g(x) = \cos x$ (inner), $f(u) = u^5$ (outer).
$(f \circ g)(x) = f(\cos x) = (\cos x)^5 = \cos^5 x$
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Let $f(u) = e^u$ and $g(x) = 4x$.
Find $(f \circ g)(x)$.
$(f \circ g)(x) = f(g(x))$: $g(x) = 4x$ (inner), $f(u) = e^u$ (outer).
$(f \circ g)(x) = f(4x) = e^{4x}$
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Functions can be composed multiple times. Let $F(u) = \ln u$, $G(v) = v^3 + 1$, and $H(x) = \tan x$.
Find the composite function $(F \circ G \circ H)(x) = F(G(H(x)))$.
This means we apply $H$ first, then $G$ to the result of $H$, and finally $F$ to the result of $G$.
Start with the innermost function: $H(x) = \tan x$.
Apply the next function $G$ to $H(x)$: $G(H(x)) = G(\tan x) = (\tan x)^3 + 1 = \tan^3 x + 1$.
Apply the outermost function $F$ to $G(H(x))$: $F(G(H(x))) = F(\tan^3 x + 1) = \ln(\tan^3 x + 1)$.
$(F \circ G \circ H)(x) = \ln(\tan^3 x + 1)$
The domain of this function requires $\tan^3 x + 1 > 0$.
Example 1. Given $h(x) = \cos(x^3)$, identify the outer and inner functions $f$ and $g$ such that $h(x) = f(g(x))$.
Answer:
To identify the outer and inner functions, we look at the sequence of operations performed on the input variable $x$.
For $h(x) = \cos(x^3)$, the input $x$ is first cubed ($x^3$), and then the cosine function is applied to the result of cubing.
Let $g(x)$ be the function applied first (the inner function), and $f(u)$ be the function applied second (the outer function), where $u$ is the output of $g(x)$.
Step 1: $x$ is cubed. This suggests the inner function is $g(x) = x^3$.
Step 2: The cosine is taken of the result from Step 1. This suggests the outer function is $f(u) = \cos u$, where $u$ represents the input to the cosine function (which is the output of $g$).
Let's check if $f(g(x))$ gives the original function $h(x)$:
$f(g(x)) = f(x^3) = \cos(x^3)$.
This matches $h(x)$.
So, the outer function is $f(u) = \cos u$ and the inner function is $g(x) = x^3$.
Chain Rule of Differentiation: Statement and Application
The Chain Rule is a fundamental rule in calculus that allows us to find the derivative of a composite function. If a function $y$ depends on a variable $u$, which in turn depends on a variable $x$, the Chain Rule gives the derivative of $y$ with respect to $x$. It is essential for differentiating many functions encountered in practice.
Statement of the Chain Rule (Leibniz Notation)
Let $y = f(u)$ be a differentiable function of $u$, and let $u = g(x)$ be a differentiable function of $x$. Then the composite function $y = f(g(x))$ is a differentiable function of $x$, and its derivative with respect to $x$ is given by the product of the derivative of $y$ with respect to $u$ and the derivative of $u$ with respect to $x$.
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
This form is intuitive as the $\frac{dy}{du}$ and $\frac{du}{dx}$ terms appear to "cancel out" the $du$ to give $\frac{dy}{dx}$. While not a rigorous proof, this serves as a helpful mnemonic.
Statement of the Chain Rule (Function Notation)
Let $g$ be a function that is differentiable at $x$, and let $f$ be a function that is differentiable at $g(x)$. Then the composite function $F(x) = (f \circ g)(x) = f(g(x))$ is differentiable at $x$, and its derivative is given by:
$F'(x) = f'(g(x)) \cdot g'(x)$
This is read as "the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function."
Intuitive Explanation ("Derivative of the Outside times Derivative of the Inside")
To apply the Chain Rule to a composite function $y = f(g(x))$, follow these steps:
- Identify the "outer" function $f$ and the "inner" function $g$. Think of $f$ as the operation applied last, and $g$ as the operation applied first to $x$.
- Differentiate the outer function $f$ with respect to its argument (the 'inside' part). This means you find $f'(u)$, but you evaluate it at $u = g(x)$. This result is $f'(g(x))$, which represents the rate of change of the outer function with respect to its composite input.
- Differentiate the inner function $g$ with respect to the variable $x$. This gives $g'(x)$, which represents the rate of change of the inner function with respect to $x$.
- Multiply the results from Step 2 and Step 3. The derivative of the composite function $y$ with respect to $x$ is the product of the two derivatives: $f'(g(x)) \cdot g'(x)$.
Think of it as peeling away layers of the function from the outside in. You differentiate the outermost layer (keeping the inner function intact) and then multiply by the derivative of the next layer inwards, and so on, if there are multiple compositions.
Application Examples
Example 1. Find the derivative of $y = \sin(x^2)$.
Answer:
The function $y = \sin(x^2)$ is a composite function. We can express it as $y = f(g(x))$.
Identify Outer and Inner Functions:
The outer function is $f(u) = \sin u$.
The inner function is $g(x) = x^2$.
Differentiate the Outer Function:
The derivative of the outer function $f(u) = \sin u$ with respect to $u$ is $f'(u) = \cos u$.
Now, evaluate $f'(u)$ at the inner function $g(x)$: $f'(g(x)) = \cos(x^2)$.
Differentiate the Inner Function:
The derivative of the inner function $g(x) = x^2$ with respect to $x$ is $g'(x) = \frac{d}{dx}(x^2) = 2x$ (using the Power Rule).
Apply the Chain Rule:
According to the Chain Rule, $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.
$\frac{dy}{dx} = (\cos(x^2)) \cdot (2x)$
[Using Chain Rule]
It is customary to write the polynomial term first:
$\frac{dy}{dx} = 2x \cos(x^2)$
The derivative of $y = \sin(x^2)$ is $2x \cos(x^2)$.
Alternate Method (Using Leibniz Notation):
Let $y = \sin u$, where $u = x^2$.
Find the derivative of $y$ with respect to $u$:
$\frac{dy}{du} = \frac{d}{du}(\sin u) = \cos u$
Find the derivative of $u$ with respect to $x$:
$\frac{du}{dx} = \frac{d}{dx}(x^2) = 2x$
Apply the Chain Rule $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$:
$\frac{dy}{dx} = (\cos u) \cdot (2x)$
Substitute back the expression for $u$ in terms of $x$, $u = x^2$:
$\frac{dy}{dx} = (\cos(x^2)) \cdot (2x) = 2x \cos(x^2)$
This confirms the result.
Example 2. Find the derivative of $y = (3x+5)^4$.
Answer:
The function $y = (3x+5)^4$ is a composite function. We can write it as $y = f(g(x))$.
Identify Outer and Inner Functions:
The outer function is $f(u) = u^4$ (raising the inner expression to the power of 4).
The inner function is $g(x) = 3x+5$ (the base of the power).
Differentiate the Outer Function:
The derivative of the outer function $f(u) = u^4$ with respect to $u$ is $f'(u) = 4u^{4-1} = 4u^3$ (using the Power Rule).
Evaluate $f'(u)$ at the inner function $g(x)$: $f'(g(x)) = 4(3x+5)^3$.
Differentiate the Inner Function:
The derivative of the inner function $g(x) = 3x+5$ with respect to $x$ is $g'(x) = \frac{d}{dx}(3x+5)$.
Using the Sum Rule and Constant Multiple Rule: $g'(x) = \frac{d}{dx}(3x) + \frac{d}{dx}(5) = 3 \frac{d}{dx}(x) + 0 = 3(1) = 3$.
Apply the Chain Rule:
Using the Chain Rule $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$:
$\frac{dy}{dx} = (4(3x+5)^3) \cdot (3)$
[Using Chain Rule]
Multiply the constant factors:
$\frac{dy}{dx} = 12(3x+5)^3$
The derivative of $y = (3x+5)^4$ is $12(3x+5)^3$.
Example 3. Find the derivative of $y = e^{\cos x}$.
Answer:
The function $y = e^{\cos x}$ is a composite function. We can write it as $y = f(g(x))$.
Identify Outer and Inner Functions:
The outer function is the exponential function: $f(u) = e^u$.
The inner function is the exponent: $g(x) = \cos x$.
Differentiate the Outer Function:
The derivative of the outer function $f(u) = e^u$ with respect to $u$ is $f'(u) = e^u$ (using the standard derivative of $e^u$).
Evaluate $f'(u)$ at the inner function $g(x)$: $f'(g(x)) = e^{\cos x}$.
Differentiate the Inner Function:
The derivative of the inner function $g(x) = \cos x$ with respect to $x$ is $g'(x) = \frac{d}{dx}(\cos x) = -\sin x$ (using the standard derivative of $\cos x$).
Apply the Chain Rule:
According to the Chain Rule, $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.
$\frac{dy}{dx} = (e^{\cos x}) \cdot (-\sin x)$
[Using Chain Rule]
Rearrange the terms:
$\frac{dy}{dx} = -\sin x \cdot e^{\cos x}$
The derivative of $y = e^{\cos x}$ is $-\sin x \cdot e^{\cos x}$.
Derivatives of Algebraic Functions using Chain Rule (Applied Maths)
In applied mathematics and general calculus, algebraic functions often appear in composite forms, particularly functions raised to a power or functions under a radical sign. The Chain Rule, when combined with the Power Rule, provides a very efficient way to differentiate such functions without resorting to tedious expansion.
The General Power Rule (Chain Rule combined with Power Rule)
A very common type of composite function is of the form where an inner function is raised to a power: $y = [g(x)]^n$. If $g(x)$ is a differentiable function of $x$ and $n$ is any real number, we can find the derivative of $y$ with respect to $x$ using the Chain Rule. This combination of rules is sometimes referred to as the General Power Rule for differentiation.
Let $y = [g(x)]^n$. We can view this as a composite function $y = f(u)$, where $f(u) = u^n$ is the outer function and $u = g(x)$ is the inner function.
According to the Chain Rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$ or $F'(x) = f'(g(x)) \cdot g'(x)$.
- Differentiate the outer function $f(u) = u^n$ with respect to $u$ using the standard Power Rule: $f'(u) = nu^{n-1}$.
- Substitute the inner function $g(x)$ back into $f'(u)$: $f'(g(x)) = n[g(x)]^{n-1}$.
- Differentiate the inner function $g(x)$ with respect to $x$: $g'(x)$.
Multiplying these results gives the derivative of $y = [g(x)]^n$:
$\frac{d}{dx}[g(x)]^n = n[g(x)]^{n-1} \cdot g'(x)$
This is a powerful formula. It tells us to differentiate the "outside" power function (bring the power down, reduce the power by one, keeping the inside function as is), and then multiply by the derivative of the "inside" function.
This rule is widely applicable to expressions like $(ax+b)^n$, $(x^2+c)^n$, $\sqrt{x^4+1}$, $\frac{1}{(x^3+2x)^5}$, etc.
Example 1. Find the derivative of $y = (5x^2 - 7x + 1)^6$.
Answer:
The function $y = (5x^2 - 7x + 1)^6$ is in the form $[g(x)]^n$, where the inner function is $g(x) = 5x^2 - 7x + 1$ and the power is $n = 6$.
We will use the General Power Rule: $\frac{dy}{dx} = n[g(x)]^{n-1} \cdot g'(x)$.
Step 1: Identify $g(x)$ and $n$.
$g(x) = 5x^2 - 7x + 1$
$n = 6$
Step 2: Find the derivative of the inner function, $g'(x)$.
$g'(x) = \frac{d}{dx}(5x^2 - 7x + 1)$
Using the Sum/Difference Rule, Constant Multiple Rule, Power Rule, and Constant Rule:
$= \frac{d}{dx}(5x^2) - \frac{d}{dx}(7x) + \frac{d}{dx}(1)$
$= 5\frac{d}{dx}(x^2) - 7\frac{d}{dx}(x) + 0$
$= 5(2x) - 7(1) + 0$
$= 10x - 7$
So, $g'(x) = 10x - 7$.
Step 3: Apply the General Power Rule formula.
$\frac{dy}{dx} = n[g(x)]^{n-1} \cdot g'(x)$
Substitute $n=6$, $g(x) = 5x^2 - 7x + 1$, and $g'(x) = 10x - 7$:
$\frac{dy}{dx} = 6[5x^2 - 7x + 1]^{6-1} \cdot (10x - 7)$
$\frac{dy}{dx} = 6(5x^2 - 7x + 1)^5 (10x - 7)$
The derivative is $6(10x - 7)(5x^2 - 7x + 1)^5$.
Example 2. Differentiate $y = \sqrt{x^3 + 4}$.
Answer:
First, rewrite the square root function using a fractional exponent:
$y = (x^3 + 4)^{1/2}$.
This is now in the form $[g(x)]^n$, where the inner function is $g(x) = x^3 + 4$ and the power is $n = 1/2$.
We will use the General Power Rule: $\frac{dy}{dx} = n[g(x)]^{n-1} \cdot g'(x)$.
Step 1: Identify $g(x)$ and $n$.
$g(x) = x^3 + 4$
$n = 1/2$
Step 2: Find the derivative of the inner function, $g'(x)$.
$g'(x) = \frac{d}{dx}(x^3 + 4)$
Using the Sum Rule, Power Rule, and Constant Rule:
$= \frac{d}{dx}(x^3) + \frac{d}{dx}(4)$
$= 3x^{3-1} + 0$
$= 3x^2$
So, $g'(x) = 3x^2$.
Step 3: Apply the General Power Rule formula.
$\frac{dy}{dx} = n[g(x)]^{n-1} \cdot g'(x)$
Substitute $n=1/2$, $g(x) = x^3 + 4$, and $g'(x) = 3x^2$:
$\frac{dy}{dx} = \frac{1}{2}[x^3 + 4]^{(1/2)-1} \cdot (3x^2)$
Calculate the new exponent: $(1/2) - 1 = -1/2$.
$\frac{dy}{dx} = \frac{1}{2}(x^3 + 4)^{-1/2} \cdot (3x^2)$
Rewrite with a positive exponent and the square root notation:
$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{(x^3 + 4)^{1/2}} \cdot 3x^2$
$\frac{dy}{dx} = \frac{3x^2}{2\sqrt{x^3 + 4}}$
The derivative is $\frac{3x^2}{2\sqrt{x^3 + 4}}$.
Differentiation of Functions using Chain Rule (General)
The Chain Rule is applicable not only to simple compositions like $f(g(x))$ but also to compositions involving multiple nested functions, such as $f(g(h(x)))$, and it can be used in conjunction with other differentiation rules like the Product Rule and Quotient Rule.
Multiple Applications of the Chain Rule (Nested Functions)
If a function is a composition of three functions, say $y = f(g(h(x)))$, we can differentiate it by applying the Chain Rule iteratively, from the outermost function to the innermost function.
Let $y = f(u)$, where $u = g(v)$, and $v = h(x)$. Applying the Chain Rule twice:
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
Now, apply the Chain Rule again to find $\frac{du}{dx}$, since $u$ is a composite function of $x$ through $v=h(x)$: $\frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx}$.
Substitute this back into the first equation:
$\frac{dy}{dx} = \frac{dy}{du} \cdot \left(\frac{du}{dv} \cdot \frac{dv}{dx}\right)$
In function notation, if $y = f(g(h(x)))$, the derivative is:
$\frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$
This pattern continues for any number of nested functions. You differentiate the outermost function (evaluating it at everything inside), then multiply by the derivative of the next layer inward (evaluating it at everything further inside), and so on, until you reach the derivative of the innermost function.
Example 1. Find the derivative of $y = \cos^3(5x^2)$.
Answer:
First, rewrite the function to clearly show the composition: $y = [\cos(5x^2)]^3$.
This is a composition of three functions. Let's identify the layers from outermost to innermost:
Outermost function: Cubing operation, $f(u) = u^3$. Its derivative is $f'(u) = 3u^2$.
Middle function: Cosine function, $g(v) = \cos v$. Its derivative is $g'(v) = -\sin v$.
Innermost function: Polynomial $h(x) = 5x^2$. Its derivative is $h'(x) = 10x$.
The composite function is $y = f(g(h(x)))$. We apply the Chain Rule $\frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$.
Step 1: Differentiate the outermost function, evaluated at the middle function.
Differentiate $u^3$ with respect to $u$, then substitute $u = g(h(x)) = \cos(5x^2)$.
Derivative is $3u^2$. Evaluating at $\cos(5x^2)$ gives $3[\cos(5x^2)]^2 = 3\cos^2(5x^2)$.
Step 2: Differentiate the middle function, evaluated at the innermost function.
Differentiate $\cos v$ with respect to $v$, then substitute $v = h(x) = 5x^2$.
Derivative is $-\sin v$. Evaluating at $5x^2$ gives $-\sin(5x^2)$.
Step 3: Differentiate the innermost function.
Differentiate $h(x) = 5x^2$ with respect to $x$.
Derivative is $10x$.
Step 4: Multiply the results from the steps.
$\frac{dy}{dx} = [3\cos^2(5x^2)] \cdot [-\sin(5x^2)] \cdot [10x]$
Multiply the constants and rearrange the terms:
$\frac{dy}{dx} = 3 \cdot (-1) \cdot 10x \cdot \cos^2(5x^2) \cdot \sin(5x^2)$
$\frac{dy}{dx} = -30x \cos^2(5x^2) \sin(5x^2)$
The derivative is $-30x \cos^2(5x^2) \sin(5x^2)$.
Example 2. Find the derivative of $y = e^{\sin(\sqrt{x})}$.
Answer:
This function involves three nested functions: an exponential function containing a sine function, which in turn contains a square root function. Let $y = f(g(h(x)))$.
Identify the Layers:
Outermost function: $f(u) = e^u$. Its derivative is $f'(u) = e^u$.
Middle function: $g(v) = \sin v$. Its derivative is $g'(v) = \cos v$.
Innermost function: $h(x) = \sqrt{x} = x^{1/2}$. Its derivative is $h'(x) = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.
Apply the Chain Rule $\frac{dy}{dx} = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$.
Step 1: Differentiate the outermost function, evaluated at the middle function.
Differentiate $e^u$ with respect to $u$, then substitute $u = g(h(x)) = \sin(\sqrt{x})$.
Derivative is $e^u$. Evaluating at $\sin(\sqrt{x})$ gives $e^{\sin(\sqrt{x})}$.
Step 2: Differentiate the middle function, evaluated at the innermost function.
Differentiate $\sin v$ with respect to $v$, then substitute $v = h(x) = \sqrt{x}$.
Derivative is $\cos v$. Evaluating at $\sqrt{x}$ gives $\cos(\sqrt{x})$.
Step 3: Differentiate the innermost function.
Differentiate $h(x) = \sqrt{x} = x^{1/2}$ with respect to $x$.
Derivative is $\frac{1}{2\sqrt{x}}$.
Step 4: Multiply the results.
$\frac{dy}{dx} = [e^{\sin(\sqrt{x})}] \cdot [\cos(\sqrt{x})] \cdot [\frac{1}{2\sqrt{x}}]$
Combine the terms:
$\frac{dy}{dx} = \frac{e^{\sin(\sqrt{x})} \cos(\sqrt{x})}{2\sqrt{x}}$
The derivative is $\frac{e^{\sin(\sqrt{x})} \cos(\sqrt{x})}{2\sqrt{x}}$.
Example 3. Differentiate $y = x^3 \ln(x^2+1)$.
Answer:
This function is a product of two functions: $u(x) = x^3$ and $v(x) = \ln(x^2+1)$. We need to use the Product Rule first. One of the factors, $v(x)$, is a composite function itself, so we will need the Chain Rule to find $v'(x)$.
Step 1: Apply the Product Rule.
The Product Rule states $\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$.
We have $u(x) = x^3$. Its derivative is $u'(x) = \frac{d}{dx}(x^3) = 3x^2$ (by Power Rule).
We have $v(x) = \ln(x^2+1)$. We need to find $v'(x)$.
Step 2: Find $v'(x)$ using the Chain Rule.
The function $v(x) = \ln(x^2+1)$ is a composite function of the form $f(g(x))$.
Outer function: $f(w) = \ln w$. Its derivative is $f'(w) = \frac{1}{w}$.
Inner function: $g(x) = x^2+1$. Its derivative is $g'(x) = \frac{d}{dx}(x^2+1) = 2x + 0 = 2x$ (by Sum and Power Rules).
Using the Chain Rule $v'(x) = f'(g(x)) \cdot g'(x)$:
$v'(x) = \left(\frac{1}{x^2+1}\right) \cdot (2x)$
[Applying Chain Rule to $v(x)$]
$v'(x) = \frac{2x}{x^2+1}$
Step 3: Substitute $u(x), u'(x), v(x), v'(x)$ into the Product Rule formula.
$\frac{dy}{dx} = u(x)v'(x) + v(x)u'(x)$
$\frac{dy}{dx} = (x^3) \left( \frac{2x}{x^2+1} \right) + (\ln(x^2+1)) (3x^2)$
[Substituting into Product Rule]
Simplify the first term:
$\frac{dy}{dx} = \frac{2x^4}{x^2+1} + 3x^2 \ln(x^2+1)$
The derivative is $\frac{2x^4}{x^2+1} + 3x^2 \ln(x^2+1)$.